I'm having trouble understand the concepts for this problem. Here's the problem:

A block is placed on a plane inclined at angle $\theta$. The coefficient of friction between the block and the plane is $\mu = \tan \theta$. The block is given a kick so that it initially moves with speed $V $horizontally along the plane (that is, in the direction perpendicular to the direction pointing straight down the plane). What is the speed of the block after a very long time?

We can see that $F_x = mg\sin \theta - mg\cos (\theta )\mu = 0$, if we gave the block an initial velocity along the plane, it should be subjected to a friction force opposite its velocity, where $f = mg \cos \theta \tan \theta = mg\sin \theta$

This however, is the answer my book gave:

The normal force from the plane is $N = mg \cos \mu$. Therefore, the friction force on the block is $N = (\tan \mu)N = mg \sin \mu$. This force acts in the direction opposite to the motion. The block also feels the gravitational force of $mg \sin \theta$ pointing down the plane. Because the magnitudes of the friction force and the gravitational force along the plane are equal, the acceleration along the direction of motion equals the negative of the acceleration in the direction down the plane. Therefore, in a small increment of time, the speed that the block loses along its direction of motion exactly equals the speed that it gains in the direction down the plane. Letting v be the speed of the block, and letting vy be the component of the velocity in the direction down the plane, we therefore have $$v + vy = C$$ where $C$ is a constant. $C$ is given by its initial value, which is $V + 0 = V$ . The final value of $C$ is $V_f + V_f = 2V_f$ (where $V_f$ is the final speed of the block), because the block is essentially moving straight down the plane after a very long time. Therefore, $2V_f = V \rightarrow V_f = \frac{V}{2}$

I don't really understand what this is trying to say. The bolded part is what's giving me problems. I think it just may be poorly written, I think I need a clearer explanation.

  • $\begingroup$ I think your textbook worded the problem poorly. $\endgroup$ Commented Aug 12, 2016 at 1:02

1 Answer 1


That's a nice problem - what book is it from? One can get the right answer quickly by projecting forces to the moving axis aligned with the instantaneous direction of V to find d/dt(V), and that is what the book seems to be suggesting (but in a rather obscure way).

Here is a different approach to the solution, probably more straightforward in terms of physics but requiring a bit of algebra and calculus, using coordinates (x,y) on the surface of the wedge; x is the horizontal axis, y is vertical (up the wedge).

There are three forces acting on the body: gravity $mg$, normal reaction $N$, and friction $R$. Since the motion is in the plane of the wedge the sum of forces projected to the normal direction is zero so $N=mg \cos{\theta}$. Within the plane there are two forces. First is the projection of gravity $mg \sin \theta$ directed downward. The second one is the friction force $R$. For $\mu=\tan(\theta)$ a body on inclined plane with the friction coefficient $\mu$ is on threshold of sliding; the standard analysis is given, e.g., in https://en.wikipedia.org/wiki/Friction. Thus the friction force is equal to $R=N \mu = mg \cos{\theta} \tan \theta = mg \sin{\theta}$ and is directed opposite to the velocity vector.

To summarize, the motion within the plane results from action of two forces of equal magnitude $mg \sin{\theta}$, one is directed down the wedge, the other is directed opposite to the velocity vector. From that one can derive that the rate of change of the total velocity magnitude $V$ is equal to that for $V_y$. Initially the body is moving horizontally with velocity $V_0$, so $V=V_0$ and $V_y=0$. Eventually the motion vector is directed down the wedge with $V_f$, so the final values are $V=V_f$ and $V_y=-V_f$. Then it follows that $V_f-V_0=-V_f$, so $V_f=V_0/2$.

The details of the calculation are shown below:

  • $\begingroup$ The main problem,perhaps,that the OP was facing is why the friction force is equal to the y component of gravity minus the normal force.it is written in your answer without any explanation. Can you please elaborate the fact.thanks. $\endgroup$
    – Pink
    Commented Dec 26, 2016 at 10:35
  • $\begingroup$ ok, added in the text of the answer, thanks for pointing out $\endgroup$ Commented Jan 2, 2017 at 2:08
  • $\begingroup$ You seem not to follow my previous comment.The real question is -why the friction force is equal in size to the y projection of the gravity minus the normal force N? which is mentioned in your second para.How did you come to that conclusion.please expand your answer on that. $\endgroup$
    – Pink
    Commented Jan 2, 2017 at 4:03
  • $\begingroup$ Ok, I see that something was not right previously in the explanations. Now it should be correct, and hopefully it is explained clearly now. $\endgroup$ Commented Jan 2, 2017 at 19:04

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